I can't proceed with the normal method to find these ones because they will intersect infinite times. Should I take an interval?

$$f(x)=\tan(x), \qquad g(x)=5x$$

Thank you.

  • 1
    $\begingroup$ Do you just have to give one intersection? Can you just say $x=0$ and not bother with the other intersections? $\endgroup$ – AHusain Oct 16 '18 at 23:46
  • $\begingroup$ You are looking for solutions of the equation $$\tan(x)=5x. $$ First, obviously, $0$ is a solution. Second, both functions are continuous ($f$ is continuous over its domain, that is, for $x\neq n\pi+\frac{\pi}{2}$ for all $n\in\mathbb{Z})$. Define $$h(x)=\tan(x)-5x, $$ so you can use the mean value theorem to approximate a solution. Third, there is several methods for solving the fixed point problem $\tan(x)=5x.$ $\endgroup$ – Mateus Rocha Oct 16 '18 at 23:58
  • $\begingroup$ What's the "normal method"? It might give some ideas for answers if the question said what you would have done if there were not infinitely many intersections. $\endgroup$ – David K Oct 17 '18 at 0:35

We can easily find the trivial intersection at $x=0$ and for $x>0$ each one of the others solutions are contained in the intervals

$$n\pi<x<\frac{\pi}2+n\pi \quad n\in\mathbb{N}$$

and can be found by numerical methods.


As gimusi answered, discrading the trivial solution $x=0$, you have a root $x_n$ such that $$n \pi < x_n < \left(n+\frac 12\right)\pi$$ Plotting the function, you noticed that the solutions are very close to the right bound.

To get rid of the discontinuities, consider instead looking for the zero's of function $$F(x)=\sin(x)-5x \cos(x)$$ You can get quite good estimates of the solution expanding $F(x)$ as a truncated Taylor series built at $x=\left(n+\frac 12\right)\pi$. Limited to $O\left(\left(x-\frac{2 n+1}{2} \pi \right)^3\right)$, this would give as the equation to solve $$9 y^2-5 (2 n+1 )\pi y+2=0\qquad \text{where} \qquad y=\left(n+\frac 12\right)\pi-x_n$$ As a result, $$x_n \approx \frac{2}{9} (2 n+1)\pi+\frac{1}{18} \sqrt{25 (2 n+1 )^2\pi^2-72}$$ Checking for a few values of $n$, $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 0 & 1.4325167 & 1.4320322 \\ 1 & 4.6695980 & 4.6695848 \\ 2 & 7.8284421 & 7.8284393 \\ 3 & 10.977358 & 10.977357 \end{array} \right)$$ For $n >3$, there is no difference at the level of $8$ significant figures.

You could have simpler estimates using the $[1,1]$ Padé approximant built at $x=\left(n+\frac 12\right)\pi$. This would give as solution $$x_n \approx \left(n+\frac 12\right)\pi-\frac{10 (2 n+1 )\pi}{25 (2 n+1 )^2\pi^2-18}$$ which is as good as the previous one.

If you need more accuracy, use Newton method starting from one of these estimates. The iterates will be given by $$x_{k+1}=x_k-\frac{\sin (x_k)-5 x_k \cos (x_k)}{5 x_k \sin (x_k)-4 \cos (x_k)}$$ For the first root, we should then have $$\left( \begin{array}{cc} k & x_k \\ 0 & 1.4325167001302000019 \\ 1 & 1.4320324138494111588 \\ 2 & 1.4320322362434419819 \\ 3 & 1.4320322362434180905 \end{array} \right)$$ which is the solution for $20$ significant figures.

  • $\begingroup$ If you look into more details, you would notice that, for large values of $n$, the distance betwenn the root and the asymptote is almost $\frac 1 {5\pi n}$ $\endgroup$ – Claude Leibovici Oct 17 '18 at 4:46

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